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我正在使用jQuery数据表通过Ajax和SQL输出数据。我可以很好地输出数据。但是,我想将两个返回对象合并到一个列中,因为它与属于同一事物有关。在一个AJAX数据表列中显示两个对象
$('#todayApt').on('show.bs.modal', function (event){
$('#todayAptList').DataTable({
"ajax": {
"url": '{{ url('panel/appointment/ajax/schedule/week/lookup') }}',
dataSrc: ''
},
"columns": [
{ "data": "date" },
{ "data": "office" },
{ "data": "block" },
{ "data": "last_name" },
{ "data": "street_1" },
{ "data": "zip_code" },
{ "data": "phone_1" },
{ "data": "service_detail" }
]
});
});
我的目标如下:
[
{
"street_1":"1234 Main St",
"phone_2":"(555) 555-5555",
"street_2":null,
"date":"2016-10-19",
"users_info_id":19,
"last_name":"Doe",
"phone_1":"(555) 555-5555",
"zip_code":90210,
"status":"scheduled",
"office":"location",
"block":"9-12",
"special_detail":null,
"mp_detail":null,
"service_detail":"Service Details"
}
]
所以service_detail,mp_detail和special_detail我想只是在标有 '详细信息' 一个单列。我可以找出如何做单列,但无法弄清楚如何做一个以上的只有一列
例预期成果:
<table id="todayAptList" class="table table-striped table-bordered dt-responsive" cellspacing="0" width="100%">
<thead>
<tr>
<th>Date</th>
<th>Office</th>
<th>Block</th>
<th>Last Name</th>
<th>Address</th>
<th>Zip Code</th>
<th>Phone</th>
<th>Services</th>
</thead>
<tbody>
<tr role="row" class="odd">
<td class="sorting_1" tabindex="0">2016-10-18</td>
<td>Location</td>
<td>3-5</td>
<td>Doe</td>
<td>1234 Main St</td>
<td>90210</td>
<td>(555) 555-5555</td>
<td>SPECIAL, MP AND SERVICE DETAILS HERE IN ONE</td>
</tr>
</tbody>
返回结果是一个JSON对象吗? – Anson
是的。我将编辑我的答案,以显示从我的SQL对象转储的正确JSON响应。 – Lynx
你能提供你期望的结果吗? – Anson